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Life of substance reduces to half at the end of one hour i.e its quantity reduces to one half of what it was at the beginning of one hour .

In how many hours , the quantity becomes less than $1$% initial quantity..

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up vote 6 down vote accepted

After $1$ hour, we have $\frac{1}{2}$ left.

After $2$ hours, we have one-half of $\frac{1}{2}$ left, so $\frac{1}{4}$.

After $3$ hours, we have one-half of $\frac{1}{4}$ left, so $\frac{1}{8}$.

After $4$ hours, we have one-half of $\frac{1}{8}$ left, so $\frac{1}{16}$.

Continue. After $6$ hours, we have $\frac{1}{64}$ left, after $7$, we have $\frac{1}{128}$. Now we are below $1\%$.

If the looked for answer is an integer, that integer is $7$. But if the decay process takes place continuously, as it probably does, the answer will be a number between $6$ and $7$. To find that number, note that if at the beginning we have an amount $A$, then after $t$ hours we have an amount $A(t)$, where $$A(t)=\frac{a}{2^t}.$$

We want the time $t$ until $A$ decays to $\frac{A}{100}$. So we have $$\frac{A}{100}=\frac{A}{2^t}.$$ This simplifies to $2^t=100$.

To solve this equation for $t$, we can use our calculator to hunt and peck our way. For example, my calculator says that $2^{6.5}\approx 90.51$, so $t\gt 6.5$. Soon you can zoom in on an excellent approximation.

Or else we can take logarithms, to any base you like. We get $\log(2^t)=\log(100)$. So $$t\log 2=\log(100),$$ and therefore $$t=\frac{\log(100)}{\log 2}.$$ I get $t\approx 6.644$.

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We use the half life formula to solve. Assume that you start with $100$% of the quantity. You want to find the time it takes to reach $1$%, so let us set up an equation.

$$ .01 = \exp(-kt). $$

Notice I have an unknown value in this equation ($k$). We can find this by using what we are given,

$$ 1/2 = \exp(-k) $$

Here we have just $\exp(-k)$ since we know $t=1\Rightarrow 1/2$ quantity left. Now use logarithms to solve for $k$ and then solve the first equation I posted.

Since $\ln(1/2)=-\ln(2)$ we have $k=\ln(2)$. Now solve

$$ .01 = \exp(-\ln(2)t). $$

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is there any simple method to solve this question ? \ – SSK Jun 13 '13 at 18:45
@SSK Not if the exact answer is called for. Many people in the physical sciences use the "rule-of-thumb" that seven half-lives reduce the initial amount of a (radioactive or biological) substance to about 1% , since $ \ ( \frac{1}{2} )^7 = \frac{1}{128} < \frac{1}{100} \ . $ (On the other hand, a lot of physicists use "e-folding times", for which the quantity falls to under 1% of its initial value in five e-folding times, since $ \ ( \frac{1}{e} )^5 \approx \frac{1}{148} < \frac{1}{100} \ . $ ) – RecklessReckoner Jun 13 '13 at 18:51
@amWhy I think a continuous process is intended: the wording in OP's statement might suggest otherwise, but I think that is just the way OP expressed the situation. – RecklessReckoner Jun 13 '13 at 18:53

Let be $n$ the initial quantity. If at each hour the quantity reduces to half of what was at the beguining of this hour, one can say that $\displaystyle \frac{n}{2^x}$ gives the quantity at the end of each $x$ hour.

If $\displaystyle \frac{n}{2^x}$ gives the quantity at each hour, then $\displaystyle n \cdot \frac{2^x}{n}=\frac{1}{2^x}$ gives the relative percentage. And so, one needs to solve $\frac{1}{2^x}\leq \frac{1}{100}$.

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